Consider a Gromov-hyperbolic group $\Gamma$ and let $\mu$ be a finitely supported probability measure on $\Gamma$. Assume that the support of $\mu$ generates $\Gamma$ as a semi-group, in other words, the random walk $X_n$ driven by $\mu$ can visit the whole group $\Gamma$.

Fact : the random walk $X_n$ almost surely converges to a point in the Gromov boundary $\partial X$ of $X$. Let $\nu$ be the exit measure on $\partial X$. Then, $(\partial X,\nu)$ is a model for the so-called Poisson boundary. The measure $\nu$ is called the harmonic measure (with respect to $\mu$).

Now consider the reverse measure $\check{\mu}$ defined by $\check{\mu}(g)=\mu(g^{-1})$ and let $\check{\nu}$ be the corresponding harmonic measure.

**Question** : Are $\nu$ and $\check{\nu}$ equivalent ?

It seems that the answer is positive. In the paper [Harmonic measures versus quasiconformal measures for hyperbolic groups][1], Blachère, Haïssinsky and Mathieu prove that $h=lv$ on a hyperbolic group if and only if the harmonic measure $\nu$ and the Patterson-Sullivan measure $\rho$ are equivalent. Here, $h$, is the asymptotic entropy, $l$ is the asymptotic drift and $v$ is the volume growth. Very roughly, the proof goes like this.

First, both measures are ergodic so they are equivalent or they are mutually singular. The Hausdorff dimension of $\nu$ is $h/l$ and the Hausdorff dimension of $\rho$ is $v$. Using the Lebesgue differentiation theorem, they show that they are equivalent if and only if they have the same Hausdorff dimension. This part is very technical, the basic ingredients being the shadow lemmas for the Patterson-Sullivan measure and for the harmonic measure.

Since the shadow lemmas for the harmonic measures $\nu$ and $\check{\nu}$ hold and since $\nu$ and $\check{\nu}$ have the same Hausdorff dimension, I think that one can adapt their proof to show that $\nu$ and $\check{\nu}$ are equivalent.

However, I'd like to prove a similar result in a broader setting and this strategy seems over-complicated and too technical. So I wonder if there exists a much simpler proof.

Note that one really needs to use hyperbolicity somewhere. For example, consider a non-centered probability measure $\mu$ on $\mathbb{Z}^d$, that is
$$p=\sum_{x\in \mathbb{Z}^d}x\mu(x)\neq 0.$$
Then, $p$ is called the drift of the random walk. By [Ney and Spitzer theorem][2], the Martin boundary of the random walk is a sphere : a sequence of points $x_n$ converges to a point in the boundary if and only if $x_n$ goes to infinity and $x_n$ converges in direction, that is $x_n/\|x_n\|$ converges. The harmonic measure is supported on the direction given by $p$. In particular, the harmonic measure associated with $\check{\nu}$ is supported on $-p$ and so $\nu$ and $\check{\nu}$ are singular.


  [1]: https://arxiv.org/abs/0806.3915
  [2]: https://www.ams.org/journals/tran/1966-121-01/S0002-9947-1966-0195151-8/S0002-9947-1966-0195151-8.pdf